Please help to make me understand why I cant optimize this function: $U=x^{1/3}*y^{2/3}$ ?

Question

If I want to maximize a production the function of which is given by $$L=-x^2+10x-2y^2+12y$$ I know I have to take the partial derivatives of of the function in respect to X and Y, so $$\frac {\partial L}{\partial X} = -2x+10 ~~\mathrm{and}~~ \frac{\partial L}{\partial Y} =-4y+12$$

Ok so now I know that if I want to find its maximum I have to equal the partial derivatives to 0 and solve . I plug the values I get $-x=5$ and $y=3-$ into the production function and I get the maximum output at 43.

Sweet, so now instead of wanting to find the maximum of this production function I want to find the maximum of an utility function which is $U=x^{1/3}y^{2/3}$, I take the partial derivatives and I get $\frac{\partial U}{\partial X} = \frac 1 3 x^{-2/3}y^{2/3}$ and $\frac{\partial U}{\partial Y} \frac 2 3x^{1/3}y^{-1/3}$.

What I do next is , as I did in the production function, equal to 0 both derivatives, but unlike before I can't find their first order condition for any of them as I have two variables in each of the derivatives. Why is that? is it because of the nature of the function?

Answer 1

In the second case there is no $(x, y)$ that makes both derivatives zero, so there is no maximum. This means that you have to look at the values for which the equations make sense, and look at the maximal value on the boundary.

This really is obvious if you look at $U = x^{1/3} y^{2/3}$, as $x$ and $y$ move away from $(0, 0)$ the value of $U$ increases monotonically.

Answer 2

I'm not sure if this answers your question, but just in case.

The function $x \mapsto x^3$ is strictly monotonic, so instead of dealing with the utility $U(x,y) = \sqrt[3]{xy^2}$, deal with $\phi(x,y) = U(x,y)^3 = xy^2$.

If you need to take derivatives, it is much easier. However, it is clear that if $x>0$, you can make $\phi(x,y)$ as large as you want by choosing $y$ appropriately. Hence it is unbounded above, and there is no maximum.

You can set the derivatives to zero (any $(x,y)$ with $y=0$ will do), but these are not global maxima (they are local maxima for $x<0$ and local minima for $x>0$, and a saddle point for $x=0$).